先對[0,n] 排序，然后從1開始，依次比較當前數字nums[i]和前一個數字nums[i-1]是否相等,如果相等，就返回當前值，即為重復數字
Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
Note:
You must not modify the array (assume the array is read only).
You must use only constant, O(1) extra space.
Your runtime complexity should be less than O(n2).
There is only one duplicate number in the array, but it could be repeated more than once.
int cmp ( const void *a , const void *b )
{
return *(int *)a - *(int *)b;
}
int findDuplicate(int* nums, int numsSize) {
int i;
qsort(nums,numsSize,sizeof(int),cmp);
for(i=1;i<numsSize;i++)
{
if(nums[i-1]==nums[i])
return nums[i];
}
return -1;
}
